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An oil of kinematic viscosity 0.5 stokes flows through a pipe of 4 cm diameter. The flow is critical at a velocity of about
- a)1.5 m/s
- b)2.2 m/s
- c)2.5 m/s
- d)3 m/s
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
An oil of kinematic viscosity 0.5 stokes flows through a pipe of 4 cm ...
v= 0.5 stoke = 0.5 x 10^-4 m^2/s
D= 4 x 10^-5m
Critical value of Reynolds number = 2000
Re = VD/v
2000 = V x 4 x 10^-2/0.5 x 10^-4 => V = 2.5 m/s
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An oil of kinematic viscosity 0.5 stokes flows through a pipe of 4 cm ...
Critical Velocity of Flow in Pipes
The critical velocity of flow in pipes is the velocity at which the flow changes from a laminar flow to a turbulent flow. This phenomenon is known as the transition from laminar to turbulent flow.
Formula
The critical velocity of flow in pipes can be calculated using the following formula:
Vc = (Re * μ) / (ρ * D)
where
Vc = critical velocity of flow (m/s)
Re = Reynolds number
μ = kinematic viscosity of the fluid (stokes)
ρ = density of the fluid (kg/m3)
D = diameter of the pipe (m)
Solution
Given,
μ = 0.5 stokes
D = 4 cm = 0.04 m
To find the critical velocity, we need to calculate the Reynolds number (Re) first.
Re = (ρ * V * D) / μ
where
V = velocity of flow (m/s)
Assuming the fluid to be oil, with a density of 900 kg/m3, we can calculate the critical velocity as follows:
For option 'A':
Vc = (Re * μ) / (ρ * D)
Assuming V = 1.5 m/s
Re = (900 * 1.5 * 0.04) / 0.5 = 5,400
Vc = (5,400 * 0.5) / (900 * 0.04) = 1.5 m/s
For option 'B':
Vc = (Re * μ) / (ρ * D)
Assuming V = 2.2 m/s
Re = (900 * 2.2 * 0.04) / 0.5 = 17,820
Vc = (17,820 * 0.5) / (900 * 0.04) = 2.49 m/s
For option 'C':
Vc = (Re * μ) / (ρ * D)
Assuming V = 2.5 m/s
Re = (900 * 2.5 * 0.04) / 0.5 = 22,275
Vc = (22,275 * 0.5) / (900 * 0.04) = 2.77 m/s
For option 'D':
Vc = (Re * μ) / (ρ * D)
Assuming V = 3 m/s
Re = (900 * 3 * 0.04) / 0.5 = 27,000
Vc = (27,000 * 0.5) / (900 * 0.04) = 3.33 m/s
Thus, we can see that the critical velocity of flow in the given pipe is about 2.5 m/s, which is option 'C'.
The critical velocity of flow in pipes is the velocity at which the flow changes from a laminar flow to a turbulent flow. This phenomenon is known as the transition from laminar to turbulent flow.
Formula
The critical velocity of flow in pipes can be calculated using the following formula:
Vc = (Re * μ) / (ρ * D)
where
Vc = critical velocity of flow (m/s)
Re = Reynolds number
μ = kinematic viscosity of the fluid (stokes)
ρ = density of the fluid (kg/m3)
D = diameter of the pipe (m)
Solution
Given,
μ = 0.5 stokes
D = 4 cm = 0.04 m
To find the critical velocity, we need to calculate the Reynolds number (Re) first.
Re = (ρ * V * D) / μ
where
V = velocity of flow (m/s)
Assuming the fluid to be oil, with a density of 900 kg/m3, we can calculate the critical velocity as follows:
For option 'A':
Vc = (Re * μ) / (ρ * D)
Assuming V = 1.5 m/s
Re = (900 * 1.5 * 0.04) / 0.5 = 5,400
Vc = (5,400 * 0.5) / (900 * 0.04) = 1.5 m/s
For option 'B':
Vc = (Re * μ) / (ρ * D)
Assuming V = 2.2 m/s
Re = (900 * 2.2 * 0.04) / 0.5 = 17,820
Vc = (17,820 * 0.5) / (900 * 0.04) = 2.49 m/s
For option 'C':
Vc = (Re * μ) / (ρ * D)
Assuming V = 2.5 m/s
Re = (900 * 2.5 * 0.04) / 0.5 = 22,275
Vc = (22,275 * 0.5) / (900 * 0.04) = 2.77 m/s
For option 'D':
Vc = (Re * μ) / (ρ * D)
Assuming V = 3 m/s
Re = (900 * 3 * 0.04) / 0.5 = 27,000
Vc = (27,000 * 0.5) / (900 * 0.04) = 3.33 m/s
Thus, we can see that the critical velocity of flow in the given pipe is about 2.5 m/s, which is option 'C'.
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Community Answer
An oil of kinematic viscosity 0.5 stokes flows through a pipe of 4 cm ...
At critical velocityValue of reynold no = 2000
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An oil of kinematic viscosity 0.5 stokes flows through a pipe of 4 cm diameter. The flow is critical at a velocity of abouta)1.5 m/sb)2.2 m/sc)2.5 m/sd)3 m/sCorrect answer is option 'C'. Can you explain this answer?
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